Solve :

11(x - 5) + 10(y - 2) + 54 = 0

7(2x - 1) + 9(3y - 1) = 25

Given,

Equations : 11(x - 5) + 10(y - 2) + 54 = 0, 7(2x - 1) + 9(3y - 1) = 25

1st equation :

⇒ 11(x - 5) + 10(y - 2) + 54 = 0

⇒ 11x - 55 + 10y - 20 + 54 = 0

⇒ 11x + 10y - 21 = 0

⇒ 11x = 21 - 10y

⇒ x = $\dfrac{21 - 10y}{11}$ ………(1)

2nd equation :

⇒ 7(2x - 1) + 9(3y - 1) = 25

⇒ 14x - 7 + 27y - 9 = 25

⇒ 14x + 27y - 16 = 25

⇒ 14x + 27y = 25 + 16

⇒ 14x + 27y = 41

Substituting value of x from equation (1) in above equation :

$\Rightarrow 14\Big(\dfrac{21 - 10y}{11}\Big) + 27y = 41 \\[1em] \Rightarrow \dfrac{294 - 140y}{11} + 27y = 41 \\[1em] \Rightarrow \dfrac{294 - 140y + 297y}{11} = 41 \\[1em] \Rightarrow 294 + 157y = 41 \times 11 \\[1em] \Rightarrow 294 + 157y = 451 \\[1em] \Rightarrow 157y = 157 \\[1em] \Rightarrow y = 1.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{21 - 10 \times 1}{11} \\[1em] = \dfrac{21 - 10}{11} \\[1em] = \dfrac{11}{11} \\[1em] = 1.$

Hence, x = 1 and y = 1.